\(\left(x+2\right)\left(x+3\right)\left(x+4\right)-x^2\left(x-5\right)=6\\ \text{⇔}\left(x^2+5x+6\right)\left(x+4\right)-x^2\left(x-5\right)=6\\ \text{⇔}x^3+4x^2+5x^2+20x+6x+24-x^3+5x^2-6=0\\ \text{⇔}14x^2+26x+18=0\)
Vì \(14x^2+26x+18>0\left(\forall x\right)\) nên phương trình trên vô nghiệm
Pt \(\Leftrightarrow\left(x^2+3x+2x+6\right)\left(x+4\right)-x^3+5x^2-6=0\)
\(\Leftrightarrow\left(x^2+5x+6\right)\left(x+4\right)-x^3+5x^2-6=0\)
\(\Leftrightarrow x^3+4x^2+5x^2+20x+6x+24-x^3+5x^2-6=0\)
\(\Leftrightarrow14x^2+26x+18=0\)
\(\Leftrightarrow14x^2+13x+13x+\dfrac{169}{14}+\dfrac{83}{14}=0\)
\(\Leftrightarrow x\left(14x+13\right)+\dfrac{13}{14}\left(14x+13\right)+\dfrac{83}{14}=0\)
\(\Leftrightarrow\left(x+\dfrac{13}{14}\right)\left(14x+13\right)+\dfrac{83}{14}=0\)
\(\Leftrightarrow\dfrac{1}{14}\left(14x+13\right)^2+\dfrac{81}{14}=0\)
Có \(\dfrac{1}{14}\left(14x+13\right)^2+\dfrac{81}{14}\ge0+\dfrac{81}{14}=\dfrac{81}{14}>0\)
\(\Rightarrow\dfrac{1}{14}\left(14x+13\right)^2+\dfrac{81}{14}=0\) vô nghiệm
Vậy pt vô nghiệm
\(=>\left(x+2\right)\left(x+3\right)\left(x+4\right)-x^2\left(x-5\right)-6=0\)
\(< =>\left(x^2+3x+2x+6\right)\left(x+4\right)-x^3+5x^2-6=0\)
\(=>x^3+4x^2+5x^2+20x+6x+24-x^3+5x^2-6=0\)
\(=>14x^2+26x+18=0\)
\(< =>2\left(7x^2+13x+9\right)=0\)
\(=>7x^2+13x+9=0\)
mà \(7x^2+13x+9=7\left(x^2+\dfrac{13}{7}x+\dfrac{9}{7}\right)=7\left(x^2+2.\dfrac{13}{14}x+\dfrac{169}{196}+\dfrac{83}{196}\right)\)
\(=7\left[\left(x+\dfrac{13}{14}\right)^2+\dfrac{83}{196}\right]=7\left(x+\dfrac{13}{14}\right)^2+\dfrac{83}{28}\ge\dfrac{83}{28}>0\)
\(=>7x^2+13x+9=0\left(vo-li\right)\)
=>pt vô nghiệm
